The length lies within the limits 6.3 ± 0.05 cm
The breadth lies within 3.5 ± 0.05 cm
The maximum possible area =6.35x3.55
=22.5425 cm^2
The minimum possible area = 6.25 x3.45
=21.5625 cm^2
The working =6.3 x3.5
=22.05 cm^2
Maximum area-working area=22.5425-22.05
=0.4925
Working area – minimum area =22.05 – 21.5625
=0.4872
Thus absolute error =( 04925 +0.4875)/2
= 0.4900
The relative error in the area is =0.49/22.05
=0.02
joseph rimiru answered the question on November 11, 2017 at 20:09
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